48 Nuclear Operators in Hilbert Spaces

Nuclear Operators in Hilbert Spaces Let E, F be two Hilbert spaces, u a continuous linear map of E into F, and J (resp. K ) the canonical antilinear isometry of E onto its dual E (resp. of F onto F ) . We call adjoint of u and denote by u* the compose of the sequence of mappings

FK-F'LEGE; u* is a continuous linear map of F into E ; it has a norm equal to the one of u (Proposition 23.3). If we denote by ( I ) and 11 11 the inner product and the norm, both in E and F, we have, for all x E E, y E F ,

(44I Y ) = (x I U * ( Y ) ) . When E = F and u = u * , the operator u is said to be self-adjoint. A mapping u E L ( E ; E ) is called positive if (u(x) I x) 0 for all x E E. A positive operator is self-adjoint: indeed, the bilinear form (u(x) Iy) is real when x = y (see p. 113). An important and well-known property of positive operators is the one stated now:

LEMMA 48.1. Let u E L ( E ; E ) be positive. There is a unique positive map v E L ( E ; E ) such thut v2 = u. Proof. By examining the coefficients of the Taylor expansion of the function C" 3 z (1 - z)1'2 ru+

about z = 0, one sees that the Taylor series converges when 1 z I = 1. From this it follows immediately that the finite Taylor series, when z has been replaced by w E L ( E ; E ) with 11 w 11 1, converge in L J E ; E), to a continuous linear map, which it is natural to denote by (I- w ) ~ / ~ . Observe that the latter commutes with any continuous linear map which commutes with w , as it is a limit (in the sense of the operators norm) is of polynomials with respect to w. For the same reason, (I-

<

488

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NUCLEAR OPERATORS IN HILBERT SPACES

self-adjoint whenever this is true of w . And of course ((I- w)1/2)2= so is w ( I - w ) =

I - w. From this we derive that, if w is positive, ((I - w)1/2) w ( ( 1 - w ) ~ / ~We ) . see therefore that

II (1- w)x 112

= ((1- w)x

I 4 - ( 4 1- w ) x I 4 < Il(1-

w ) x II II

which proves that, in the case where w is positive and have 11 Z - w 11 1. But then we may consider

<

1) w 1)

*

111

< 1, we

( I - (I - w))1’2,

which we shall denote precisely by K w . Now, if 11 u I( is arbitrary, (11 u ll-lu)l/z. This proves the existence of the we set dii = I( u mapping v (now denoted dG) of Lemma 48.1. Suppose there was a second positive map w E L2(E;E) such that w2 = u. Then w commutes with u and therefore with du, in view of the earlier considerations. This implies that we have o=w2-

(dU)Z= (w - h ) ( w + di).

We apply the right-hand side to a vector x E E and conclude that fi on the image of ( w 4 6 ) ; the orthogonal of this image is the kernel of (w qii)(as this operator is self-adjoint), which is the intersection of Ker w with Ker fi,as w and dii are positive; and on Q.E.D. this intersection, we have trivially w = 2/U. w =

+

+

COROLLARY. Let u be a positive operator. If x 0, we have u ( x ) = 0.

( ~ ( x )I x) =

E

E satisjies the equation

Let v be an operator such that v*v = u; then ( ~ ( x1 )x) = v ( x ) = 0 implies u ( x ) -=v * ( v ( x ) ) = 0. T h e unique positive-operator v of Lemma 48.1 is often denoted by 4;and called the positive square root of u. Let us now consider an arbitrary continuous linear map of a Hilbert space E into another Hilbert space I;; let u* be the adjoint of u. Then u* u is a positive operator of E into E; let us denote by R its positive square root. We have 1) R(x) 11 = 11 ~ ( x11) for all X E E and therefore Ker R = K e r u. Let us then define the following continuous linear map U : E + F , U(x) = u ( x l ) if x = R(xl)E Im R ; Proof.

11 v ( x ) [I2 and

-

then U is extended by continuity to the closure of I m R ; U(x) = 0

if

XE

Ker R.

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[Part I11

We have, for all x E E,

I1 U(R(X))ll = II u ( 4 l = II R(.)l.

This means that U is an isometry of Im R onto Im u. We have u=UOR.

One refers often to R as the absolute value of u. Our purpose is to study the operators u : E +F which are nuclear and to characterize them. But, because of the isometric properties of the operator U above, it is clear that u will be nuclear if and only if its absolute value R is nuclear. I n other words, it suffices to study the nuclear operators which are positive mappings of a Hilbert space E into itself. But we know that nuclear mappings are compact: we may therefore restrict our attention to compact positive operators. These have simple and beautiful spectral properties, discovered at the beginning of this century by Fredholm and F. Riesz. For the benefit of the student who does not have a treatise on Hilbert space theory within reach, we recall the statement and the proof of the main theorem on compact positive operators.

THEOREM 48.1. Let E be a Hilbert space, and u a positive compact operator of E into itself. T h e is a sequence of positive numbers, decreasing and either +te or converging to zero, A,

> A, >

*--

> A, >

-*a,

v k of E (k = 1, 2,...) with the following properties: (1) the subspaces v k are pairwise orthogod; (2) for each k and all x E V, , ~ ( x = ) A$; ( 3 ) the orthogonal of the subspace spanned by the union of the V, is equal to the kernel of u.

and a sequence of nonzerofinite dimensional subspaces

Proof. Let t be the supremum, on the unit sphere of E, of the nonnegative ;function (u(x) I x). We use the fact that the closed unit ball of E is weakly compact (Proposition 34.1); there is a weakly converging sequence {xu}, 11 xu 11 = 1, such that (u(xJ I x.) -+t . But as u itself is a compact operator, we may suppose that the sequence (u(xu)} converges in the sense of the operators norm. This implies at once that, if x is the (weak) limit of the x u , (u(x) 1 x) is equal to the limit of the (u(xJ I x"), i.e., to t. Thus we have ((t1 - ) .(u 1 x) = 0.

But, because of our choice of t , tI - u is positive. Now if v is positive

Chap. 48-43

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49 1

and if ( ~ ( x )I x) = 0, we have v(x) = 0. Indeed, ( ~ ( x )I x) = (1 1/6(x) and &(x) = 0 implies (+)2(x) = 0. Thus we have u(x) = tx.

Let us denote by V , the linear subspace of elements x E E such that B, be the unit ball of V,, i.e., the intersection of the unit ball of E with V , . By hypothesis, u(Bl)is precompact; but it is equal to tB, , hence is closed and B, must be compact. Thus V , is finite dimensional. From now on we write A, instead oft. If Vk is the orthogonal of V , , we have u( V t )C V i as u is selfadjoint. Thus, by restriction, u defines a continuous linear map of Vf into itself which is clearly compact and positive. We may repeat the above procedure with E replaced by V f . The maximum h, of the function ( ~ ( x )I x) on the unit sphere of Vf is < A , , for otherwise there would be an element x in this unit sphere such that ~ ( x = ) A,x and x would belong to V , . We then build the sequences { h k } and {Vk} by induction on k, taking v k = {x E E; u(x) = A g } . If the procedure comes to a halt after a finite number of steps, say K steps, it means that the maximum of the function (u(x) I x) on the orthogonal of V , f V k is equal to zero; this orthogonal must then be the kernel of u (note that in this case the image of u is finite dimensional). If the procedure does not stop after a finite number of steps, the decreasing sequence {Ak} must converge to zero. Otherwise we would be able to find an orthonormal sequence of vectors x, such that u(xJ = &,xV with 2 c > 0. But as u is compact, the sequence {u(x.)) should contain a converging subsequence, which is absurd as u(x) = tx. Let

+

Finally, if an element x E E is orthogonal to all ‘the v k , we must have, in view of their definition, (u(x) 1 x) = 0, hence u(x) = 0. Q.E.D. We recall the well-known terminology; the numbers are the e&nva~#es of u; v k is the eagerupace of u corresponding to the eigenvdue Ak ; dim V k is sometimes called the muZt+Zicity of the eigenvalue Ak. The sum of the series m

hkdhv

k

k=l

is called the trace of u and denoted by T r u. For each k, let P k be the orthogonal projection of E onto Vk (see p. 120). The finite sums

492

p a r t I11

TENSOR PRODUCTS. KERNELS

converge to u for the operators norm as K 4a.We may write (48.1)

This representation of u is called the spectral decomposition of u. In the preceding notation, which makes use of infinite series, it is to be understood that, if the image of u is finite dimensional, the series in question are finite. We may now state and prove the main theorem on the subject of nuclear operators in Hilbert spaces:

THEOREM 48.2. Let E, F be two Hilbert spaces, linear map, and R its absolute value.

u :E

+.F a

continuous

The following properties are equivalent:

(a) (b) (c) If u

u is nuclear;

R is nuclear; R is compact and T r R is finite. is nuclear, Tr R is equal to

)I u )ITr

.

Proof. As we have already pointed out, we have u = U o R, where U is an isometry of R(E) onto u(E) C F,and it suffices to prove the equivalence of (b) and (c) and the fact that, if R is nuclear, T r R is equal to (1 R llTr . In other words, we may as well suppose that u E L(E; E ) and that u is positive. Suppose that u is nuclear; then u is compact (Proposition 47.3) and has therefore a spectral decomposition

Let us consider the finite sums

+ + PK,we see

they converge in norm to u. If we set QK = PI that we have u, = Q K u . Then (Proposition 47.1)

11 uK IITr < 11 P K 11 11

IlTr =

11 ?A Ill7 -

On the other hand, the trace form (see p. 485) is a continuous linear

Chap. 48-61

NUCLEAR OPERATORS I N HILBERT SPACES

a,,

493

form on E' E of norm one. Using the fact that, if a map u : E +. E is defined by an element 8 E E' @ E, we have 11 u / I T r = 11 8 lls , we have

I T~(uK)I < II U K llTr < II

llTr

.

But as we have pointed out, when dealing with a linear map defined by an element 8 E E' @ E, the trace form is equal to the trace in the usual sense, therefore K

dim

k=l

v k

< 11

IITr

9

+

00, we see that the where we have set V , = P,(E). By taking K - + trace of u is finite, which proves (c). Conversely, suppose that u is compact and that T r u is finite. By using the spectral decomposition of u, (48.1), we see that

and the proof of Theorem 48.2 will be complete if we show that

11 pk

IITr

< dim

vk

If we select an orthonormal basis e l , Pk =

3

v k = Pk(E)*

..., er in V , , we can write

r

2ei 0e, ,

i=l

where e; is the linear form x view of Proposition 47.2,

11 Pk

-+

(x I ei) on E. We have, therefore, in

r

< 1 1 = Tr(PK]l=dim v k .

Q.E.D.

i=l

Theorem 48.2 provides a motivation of sorts for the name trace-norm. Observe that, if u is a positive nuclear map of a Hilbert space E into itself, its trace, as an operator, is equal to the trace of an element 8 E E' @; E representing it. As a matter of fact, one can show that such an element 8 is unique, in other words, that the canonical mapping of E @, E onto L1:F; E ) is injective; it follows then from the definition of the trace-norm that it is an isometry. We leave the proof of the following corollary of Theorem 48.2 to the student:

-

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COROLLARY. A continuous linear map u : E F of Hilbert spaces is nuclear if and only if there are two orthogonal sequences { x k } , {yk}in E and F,respectively, and a sequence { h k } in P such that :

" )

1

I Xk)Yk

k

a

We consider now the space E @.,F ( E and F are Hilbert spaces). We recall that the topology E on E @ F is induced by &(E: ;F). The equicontinuous subsets of E' are identical to the bounded sets and, as E is reflexive, its weakly compact subsets are identical to its bounded subsets; in other words, L,(E; ;F ) = L , ( E ; F), the space of bounded linear operators E +F equipped with the operators norm (E carries its dual norm). As for E @=F, it is the closure, in the sense of the operators norm, of the continuous linear mappings whose image is finite dimensional.

THEOREM 48.3. Let E and F be Hilbert spaces; E the space of compact operators of E into F.

BEF

is identical to

Proof. E g 8 F is contained in the set of compact operators; indeed every continuous linear map with finite dimensional image is obviously compact. On the other hand, we have the general result: 48.2. Let E, F be two Banach spaces. The set of compact linear LEMMA operators of E into F is closed in L,(E; F ) . Proof. Let u : E +F be the limit (for the operators norm) of a sequence of compact operators. Let E be > 0 arbitrary; let us denote by B the closed unit ball of E, by B, the one o f F . There is a compact operator v : E +F such that (u - v ) ( B )E eB1 . There is a finite set of points x1 ,..., x, in B such that a(B) C (v(x1)

whence U(B)

c

(IL(X1)

+ cB1)

U *.* U (v(xr)

+ 3&&) u

'**

u (U(Xr)

+ eB1)

+ kB1).

This proves the lemma and therefore that every element of E B s F is a compact operator. In order to see that every compact operator u : E + F belongs to E BE F, it suffices to observe that u is the limit, for the operators norm, of continuous linear mappings with a finite dimensional image. Indeed, we write u = U o R, with R : E + E'

Chap. 48-81

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495

compact positive and U,an isometry of R(E’) into F. We then use the spectral decomposition of R ,

R

=

C h$k

(Cf.

(48.1)).

k

The sequence of mappings

converges to u for the norm; but Pk maps E‘ onto a finite dimensional Q.E.D. subspace, hence u 0 Pk(E) is finite dimensional. Let us take a look now at the space B(E‘,F‘), the space of continuous bilinear forms on E’ x F equipped with its natural norm, the supremum of the absolute value on the product B’ x B; of the unit ball of E and the unit ball of F . If u : E ’ + F , we can associate with u the bilinear form on E x F’, 22 : (XI, y‘)

-

(y’, U(X‘)>,

which is obviously continuous. Conversely, let 4 E B(E’,F’); then u+ : x’ (y‘ $(x’, y’)) is a continuous linear map of E into F” = F; we see immediately that 4 = zZ4 . All this means that B ( E ,F‘) is canonically isomorphic to L(E’;F ) ; it is evident that the isomorphism extends to the norms. Thus we have the natural mappings ‘c-t

‘c-t

E &F+LyE’;F)-+E

&F+L(E;F)e

B(E,F’).

The first space carries the r-norm, the second one the trace-norm, and the last two the operators norm. All the mappings are continuous; the last one is an isometry (into!). It will follow, from what we are going to say now, that the first mapping is an isometry (onto!). For this, we study the duals of the spaces above. We know what the dual of the first one is: B ( E , F ) r L ( E ; F ’ ) .By transposing the mapping (with dense image) E @, F + E Be F, we obtain an injection of the dual of E BeF into B(E, F). Its image is denoted (in general) by J(E, F ) ; the elements of J(E, F ) are called integralform on E x F (see next chapter, Definition 49.1; this definition is hardly justified in the case of Hilbert spaces, as we shall soon see). At any event; the elements of J(E, F) can be identified to certain continuous linear mappings of E into F‘. What are they ? Before we can answer this question, we must recall the duality

496

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bracket between E @, F and L(E;F’). Let 0 be an element of E @, F, and u an element of L ( E ; F ‘ ) . We may consider the extended tensor product IZidentityofF. u &,I: E @,,F+F’ &,F, Then ( u @,,I)(0) is an element of F‘ @, F; we may consider its trace (see p. 486). We have ( u , 0) = Tr((u

B,,W)).

If we use a representation of 0 of the form m

hkxk@yk k=O

+

with CkI Ak I < co, {xk} and Cyk} being two sequences converging to zero in E and F, respectively, we have (48.2)

At this stage, we may show that we need not distinguish all the time between E @*F and L1(E‘;F ) . Since the trace-norm is the quotient of the n-norm modulo the kernel of the canonical mapping E @, F + L1(F’;E), it suffices to show that this mapping is one-to-one. This is a consequence of the following result:

THEOREM 48.4. If E and F are Hilbert spaces, the canonical mapping of E @, F into E @=F is one-to-one. Proof. It suffices to show that the transpose of the mapping in question has a weakly dense image; since the mapping itself has trivially a dense image, its transpose is one-to-one. In other words, we must show that J(E, F) (E @=F)’ is weakly dense in B(E, F ) g ( E @, F)’; here, weakly has to be understood in the sense of the duality between B(E, F ) and E @,,F. We identify B(E, F) to L ( E ;F ) and we note that the set of w : E -+F’ with finite dimensional images (and which moreover are linear continuous!) belong clearly .to J(E, F ) . It will therefore suffice to show that, if 8 E E @, F, to every u E L ( E ;F ) there is such a w with the property that I ( u - o,8) I < E ( E > 0 arbitrary). Suppose then that we have $roved the following fact: (48.3)

To every compact subset K of E and every neighborhood V of 0 F’, there is a continuous linear mapping w : E +F’ with finite dimensional image such that (24

- w)(K)C

v.

Chap. 48-10]

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NUCLEAR OPERATORS IN HILBERT SPACES

We shall combine (48.3) with Eq. (48.2): we shall take, as set K, the set

{ x ~ ) ~ - , , ~ .u . . ,(0) and, as set V, the multiple pSo of the polar So of the sequence S = {Yk), with p = I I). We have then I ( u - v , 8) 1

<

e/(zk

e. Now, Property (48.3) follows easily from the lemma:

LEMMA 48.3. Let H be a Hilbert space. Then, to every compact subset C of H and to every nezghborhood of zero V in H , there is a continuous linear map w : H + H with finite dimensional image such that, for all x

E

c,

x

-

w(x) E

v.

Before proving the lemma, let us show how it enables us to complete the proof of Theorem 48.4. We apply it with H = F' and C = u(K), where K is an arbitrary compact subset of E. Let us choose w as in Lemma 48.3 and se't v = w o u ; the image of v is finite dimensional, and (I'- w ) ( C ) C V(I':identity of F') is equivalent with (u - v ) ( K )C V , whence (48.3).

Proof of Lemma 48.3. We may assume that V is a closed ball centered at the origin; then C can be covered by a finite number of balls of the form x, + V; let M be the (finite dimensional) linear subspace of H spanned by their centers. We have C C M V; if w is the orthogonal projection of H onto M , the norm of x - w ( x ) is necessarily radius of V, since there is a point x" E M such that x - x" E V ;thus x - w ( x ) E V .

+

<

Let u E L ( E ;F') B ( E , F ) , and 8 E L1(E';F ) E @, F. The transpose of u, fu, is a continuous linear operator of F into E ; the compose 8 o Iu maps F into itself. It is a nuclear operator, by Proposition 47.1. As one sees immediately,
THEOREM 48.5. If E and F are HiIbert spaces, the dual of E canonically isomorphic to L1(E ; F').

F is

The isomorphism extends to the norms: E @=F carries the operators norm (which is equal to the e-norm); L1(E;F'),the space of nuclear operators E -+F', carries the trace-norm (which is equal to the vnorm). Theorem 48.5 is due to J. Dixmier and R. Schatten. If we use the canonical antilinear isometry of a Hilbert space onto its dual, we can give the following more striking statement of Theorem 48.5:

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[Part 111

THEOREM 48.5’. Let E and F be Hilbert spaces. The dual of the space of compact linear operators E+F, equipped with the operators norm, is the space of nuclear operators E - F , equ@ped with the trace-norm, and its bidual is the space L,(E; F ) of all continuous linear operators E +F.

-

Proof of Theorem 48.5. As we have said, we must show that, if 8 (u, 8) is continuous on E &F, then u must be nuclear. Let us write tu =

UR

(we omit the “compose” sign 0).

where R is the absolute value of *U and U is an isometry of R(F) into E . We shall take advantage of the following result of the spectral theory of linear operators in Hilbert spaces: ;f R is a positiwe bounded operator ( o fF into itself), which is not compact, there exists a bounded linear operator G such that GR is the orthogonal projection P onto a closed subspace of injinite dimension. We then choose

e =w

~

~

-

(:l E

jq,

where w is a continuous linear map of F into itself having a finite dimensional image. We have then (u, 0 )

= Tr(8‘u) = Tr(wP).

Now, if w remains in the unit ball of L,(F, F),we have

II 0 II < 1I CJ II II GI1 < II GII, so that the norm of 8 (as an operator!) remains bounded. But the e-norm

is precisely the operators norm. On the other hand, if w is the orthogonal projection of F onto a linear subspace of P(F)of dimension n, we have W P = w and Tr(w) = n. Since dim P(F)= 00, we may take n + +00. In this way, we see that I ( u , 8> I does not remain bounded, although 8 remains bounded in the &-topology.We have reached a contradiction. It means that R, and therefore b, must be compact. But then we may write (cf. (48.1))

+

00

R=

hkPk, k-1

where the Pk are orthogonal projections on (pairwise orthogonal) finite dimensional subspaces v k of F. We choose now 8 = WU-’, whence Pk, whence Tr(8‘u) = Tr(wR), and v =

xf=l

Tr(wR)

N

=

hk k-1

dim

v k .

Chap. 48-12]

NUCLEAR OPERATORS IN HILBERT SPACES

On the other hand, Thus we have

I ( u , 8) 1

< const

499

(1 8 (1 (11 11: operators norm).

C X k dim v k < const (1 e (1. N

k=l

By going to the limit N + co, we see that T r R is finite and is at most equal to the norm of the linear form 8
-..

But on the other hand, if u is nuclear, so is % and 8 o Iu. We have (Proposition 47.1; here 8 is any element of L(E';F))

I<%e>l = 1 Tr(8 0 ")l

< 11

0 tY llTr

< 11

I/ 11 tU (IT1

and the proof is complete if we apply the corollary of Proposition 47.5 (in relation to the last estimate, we recall that the trace-form is a continuous linear functional on F @- F L1(F;F) of norm one).

=